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1 Chapter 15: Mersenne Primes and Perfect Numbers Practice HW p. 108 # 1, 2, 3a, 3b In this section, we examine how to use Mersenne primes to generate perfect numbers. We start with the following definition. Definition 15.1: A number n is a perfect number if the sum of its proper divisors (all of the divisors of n less than n excluding n itself) equals n. Example 1: Explain why 6 and 28 are perfect numbers. Solution: █ Example 2: Explain why 15 is not a perfect number. Solution: █ This leads to the next question: How do we generate larger perfect numbers? The next theorem answers this question. 2 Theorem 15.1: Euclid’s Perfect Number Formula. For a prime p, if 2 p 1 is a prime number (a Mersenne prime), then a perfect number can be generated using the formula 2 p 1 (2 p 1) Proof: █ 3 Example 3: Use the prime p 5 to generate a perfect number and demonstrate that the result is a sum of its proper divisors. Solution: █ Example 4: Use the prime p 7 to generate a perfect number. Solution: █ 4 Sigma Function Definition 15.2: Given a number n, the sigma function of n, denoted as (n ) , is the sum of all of the divisors of n, including n itself. Example 5: Compute ( 6) Solution: █ Example 6: Compute ( 20) Solution: █ Facts about the Sigma Function 1. If n is a perfect number, then (n ) 2n or n (n ) 2 . 5 2. If p is a prime and k 1 , then ( pk ) 1 p p2 p3 pk p k 1 1 p 1 Proof of Fact 2: █ 3. If gcd( m, n ) 1 , then ( m n ) (m) ( n ) . Proof: Exercise 1, p. 102 6 Example 7: Compute (72501) Solution: █ Example 8: Compute ( 2882400500) Solution: █ The next question we want to ask is there any other way to generate a perfect number other than using Euclid’s Perfect Number Formula. The next theorem answers this question and shows what form all perfect numbers have. 7 Theorem 15.2: Euler’s Perfect Number Theorem. If n is an even perfect number, then there is a Mersenne prime of the form 2 p 1 that n can be written in the form n 2 p 1 (2 p 1) Proof: If n is an even number, then n has the form n 2 k s , where s is an odd integer and k 1. Now, (n) (2k s) (2k ) ( s) (note that gcd( 2 k , s ) 1) 2 k 1 1 ( s) 2 1 (use Fact 2 above) ( 2 k 1 1) ( s ) (simplify) Also, since n is a perfect number, we have (n) 2n 2(2k s) 2k 1 s Hence, we have (n) (2k 1 1) ( s) and (n) 2k 1 s . Equating these two results gives * (2k 1 1) ( s) 2k 1 s By divisibility, this implies that 2k 1 | [( 2k 1 1) ( s)] . Since gcd( 2k 1,2k 1 1) 1 , this implies that 2k 1 | ( s) or that ( s) c 2k 1 . Substituting ( s) c 2k 1 into * gives (2k 1 1) c 2k 1 2k 1 s or s (2k 1 1) c Thus, we have established that s (2k 1 1) c and ( s) c 2k 1 . (Continued on Next Page) 8 We want to next show c 1 . If c 1 , then c | s and s has the divisors 1, c, and m (plus probably others). Since, (s ) = the sum of all of the divisors of s, we have ( s) 1 c s 1 c ( 2 k 1 1 ) c (note that s ( 2 k 1 1) from above) 1 c 2 k 1 c c (Distribut e) 1 2 k 1 c (Cancel c' s) 1 ( s) (Substitut e ( s ) 2 k 1 c from above) Thus, we have established that ( s ) 1 ( s ) which is impossible. Thus, c 1 and s 2 k 1 1 . This implies that 2 k 1 s 1 and hence ( s) 2k 1 s 1 . If ( s ) s 1 , then the only divisors of s are 1 and itself. Thus, s must be prime. The only way s 2 k 1 1 can be prime (and hence a Mersenne prime) is if its exponent k 1 is prime. Setting k 1 p , where p is prime, we have k p 1 . Thus n 2k s 2k (2k 1 1) 2 p 1 (2 p 1) . This completes the proof. █ Fact: The only perfect numbers that have ever been found are even. There have never been discovered any odd numbers that are perfect. It may be true that there are not any more but nobody has ever formally proven this result.

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